Question

What is the formation reaction of aluminum chloride molecule and its enthalpy of formation?

Answer 1

You may also mean the liquid or the gas, but I assume you mean the solid. Thus, I assume you mean the process that produces `"AlCl"_3(s)` from the elemental states of the atoms.

Well...

• Chlorine naturally exists as a diatomic gas, `"Cl"_2(g)` at `25^@ "C"` and `"1 atm"`.
• Aluminum naturally exists at `25^@ "C"` and `"1 atm"` as a solid.

Therefore, the unbalanced standard formation reaction is:

`"Cl"_2(g) + "Al"(s) -> "AlCl"_3(s)`

Balancing it to ensure that we form one mol of product (as it is defined for a standard formation reaction), we write:

`color(blue)(barul(|stackrel(" ")(" "3/2 "Cl"_2(g) + "Al"(s) -> "AlCl"_3(s)" ")|))`

And this has an enthalpy of formation of `DeltaH_f^@ = -"705.63 kJ/mol"`.

Keep in mind that this compound is not much of a molecule; its electronegativity difference is

`EN_(Cl) - EN_(Al) = 3.16 - 1.61 = 1.55`,

which indicates at least moderate ionic character in the bonds, despite the trigonal planar symmetry making this compound nonpolar. This is reflected somewhat in its melting point of `192.4^@ "C"`, showing it is a solid at room temperature and pressure.